Mathematics 3 Analysis & Integration 2004 Notes on Integration 1. Null sets A subset E of R is null if, given ε > 0, we can find a countable collection of open ∞ ∞ S P intervals {Ij} such that E ⊆ Ij and |Ij| < ε. (It’s easy to see the word ‘open’ can be j=1 j=1 omitted.) Clearly singletons are null. So is any countable set. Subsets of null sets are null. Notation: For an interval I of the form I = [a, b], (a, b), (a, b] or [a, b), |I| := b − a denotes its length. If E is a finite union of disjoint bounded intervals, E = I1 ∪ .... ∪ IN , then N P |E| := |Ij| . It is tedious but routine to show that this is well-defined, and, as a consequence, j=1 if E1 and E2 are such sets with E1 ⊆ E2, then |E1| ≤ |E2|. ∞ S Theorem 1.1 If Ej (j = 1, 2, ....∞) is null, so is Ej. j=1 j+1 Proof Let ε > 0; cover Ej by intervals of total length < ε/2 . The union of all such ∞ P −j−1 intervals has length ≤ ε 2 = ε/2 < ε. j=1 However not every null set is countable: Example: The Cantor middle third set 1 2 Let E0 = [0, 1]; let E1 = 0, 3 ∪ 3 , 1 , the set obtained by removing the middle third of 1 2 1 2 7 8 E0; let E2 = 0, 9 ∪ 9 , 3 ∪ 3 , 9 ∪ 9 , 1 , the set obtained by removing the middle third of each of the two intervals comprising E1. Continuing in this way we obtain Ej, a union of j −j 2 intervals of length 3 , and Ej+1 is obtained by removing the middle third of each of these ∞ T j intervals. Let E = Ej. As the intervals comprising Ej have total length (2/3) which tends j=0 to 0 as j → ∞,E is null. E is uncountable by the usual Cantor argument. Example: Generalised Cantor sets 1 Consider the unit interval I = [0, 1]. Let F1 be the “middle” open subinterval of length 5 . Let 1 F2 be the union of the 2 “middle” open subintervals of I \F1, each of length 52 . Having defined j F1,F2, ...., Fj such that I \ (F1 ∪ .... ∪ Fj) consists of 2 closed intervals, we define Fj+1 as j 1 the union of the 2 “middle” open subintervals of I \ (F1 ∪ .... ∪ Fj), each of length 5j+1 . An ∞ S example of a generalised Cantor set is given by E = I \ Fj . Notice that E contains no j=1 j−1 j nontrivial interval, and that |Fj| = 2 /5 , so that ∞ ∞ j X 1 X 2 1 2/5 1 |F | = = = . j 2 5 2 (1 − 2/5) 3 j=1 j=1 We shall soon see that this shows that E is not null. (One can play games by varying the ratio 1/5.) 1 ∞ ∞ P Theorem 1.2 If [a, b] is covered by open intervals {Ij}j=1, then |Ij| ≥ b−a. In particular, j=1 [a, b] is not null, if a < b. ∞ P Proof Assume that |Ij| < b − a. By the Heine-Borel Theorem, we may find a finite j=1 N S subcollection {I1, ....IN } with [a, b] ⊆ Ij. Then, by the remark preceding Theorem 1.1, j=1 N N S P b − a ≤ | Ij| ≤ |Ij| , which is a contradiction. (The final inequality while ”obvious”, can j=1 j=1 easily be proved by induction on N.) Theorem 1.3 Let I be a bounded interval. Suppose that Fj (j = 1, 2, ....∞) is a finite ∞ S union of disjoint intervals, Fj ∩ Fk = ∅ when j 6= k, each Fj ⊆ I and that I \ Fj is null. j=1 ∞ P Then |Fj| = |I| . j=1 N N S P Proof Since Fj ⊆ I for all N we have |Fj| ≤ |I| for all N (using disjointness of the j=1 j=1 ∞ ∞ P P {Fj}) and so |Fj| ≤ |I| . Thus we have to show |Fj| ≥ |I| . Without loss of generality j=1 j=1 we may assume that I is closed and that each Fj consists of open intervals as this affects ∞ S neither |Fj|, |I| nor the statement that I \ Fj is null. (We have modified matters only on a j=1 ∞ ∞ P S countable, hence null set.) Suppose for a contradiction that |Fj| < |I| . Cover I \ Fj by j=1 j=1 ∞ ∞ ∞ P P ∞ ∞ open intervals {Ji}i=1 with |Ji| < |I| − |Fj| . Then {Fj}j=1 ∪ {Ji}i=1 gives a cover of i=1 j=1 I by open intervals of total length less than |I| , in contradiction to Theorem 1.2. Corollary The generalised Cantor set constructed above is not null – if it were we’d have ∞ ∞ P P 1 to have |Fj| = 1, and we showed that |Fj| = . j=1 j=1 3 Null sets will be systematically regarded as ‘negligible’ in integration theory. A property of the real numbers holds almost everywhere or holds for almost all x if it holds for all real numbers except those in some null set. Thus, for example, χ = 0 almost everywhere. Q 1 x ∈ E Notation: if E ⊆ , χ (x) = R E 0 x∈ / E. 2. Integration – what are we aiming for? We wish to develop an integral with the following features: R (i) if f : R → R is “nice”, then f should represent the “area under the graph of f” 2 (ii) if f ≥ 0, R f ≥ 0 (iii) the integral should be linear ∞ P (iv) if Ij (j = 1....∞) is a sequence of pairwise disjoint intervals with |Ij| < ∞, then j=1 R ∞ ∞ ∞ P χ S should be integrable and χ S should equal |Ij| . Ij Ij j=1 j=1 j=1 Of course we also wish the integral to be calculable by the standard techniques of integral calculus (antiderivatves, parts, substitution ....) for sufficiently nice integrands. n P If φ = cjχIj is a step function (i.e. a finite linear combination of characteristic functions j=1 of bounded intervals) then this wish list prescribes that we must have n Z X φ = cj |Ij| . j=1 We will then use analysis to extend the definition of the integral to a wider class of functions. Convention All functions f have domain R, and usually have codomain R, (but we will in occasion allow f to take the values ±∞). If g :[a, b] → R, we extend g to be zero outside [a, b] to obtain a function with domain R. (We make an exception to this convention in Sections 7-9.) 3. Integration of Step functions Definition A step function φ : R → R is a finite linear combination of characteristic functions of bounded intervals, i.e. n X φ = cjχIj j=1 where |Ij| < ∞. Evidently, φ is a step function if and only if ∃x0 < x1 < ... < xN such that (i) φ(x) = 0 for x < x0 and x > xN (ii) φ is constant on (xj−1, xj) 1 ≤ j ≤ N. We say that such a φ is a step function with respect to {x0, ..., xN }. We define the integral of such a φ by N Z X φ := φj(xj − xj−1) j=1 where φj is the constant value of φ on (xj−1, xj). Note that if φ is a step function with respect to {x0, ..., xN }, and xj−1 < c < xj, φ is also a step function with respect to R {x0, ..., xj−1, c, xj, ..., xN } and the two definitions of φ agree. Thus if φ is a step function with respect to {x0, ..., xN } and also with respect to {y0, ..., yM }, upon ordering {x0, ..., xN } ∪ 3 R {y0, ..., yM } as z0 < .... < zK (K ≤ M + N) we see that the definitions of φ with respect to {x0, ..., xN }, {z0, ..., zK } and {y0, ..., yM } all agree. Thus R φ is well-defined for any step function φ. It is clear that if φ and ψ are step functions and α, β ∈ R, then αφ + βψ is a step function; moreover Z Z Z (αφ + βψ) = α φ + β ψ (*) as if we list all the “jump points” of either φ or ψ together as {x0 < .... < xN }, the left hand N N N P P P R R side is (αφj + βψj)(xj − xj−1) = α φj(xj − xj−1) + β ψj(xj − xj−1) = α φ + β ψ. j=1 j=1 j=1 Similarly if φ ≥ ψ, with φ, ψ step functions, then R φ ≥ R ψ. R Since χI = |I| , linearity of the integral (*) implies that n n Z X X cjχIj = cj |Ij| . j=1 j=1 So points (i)-(iii) of our “wish list” are adequately resolved for step functions. How about point (iv)? As so far we’ve only defined the integral for step functions, (iv) is tantamount to asking: ∞ S If {Ij}j=1 is a disjoint sequence of intervals with union Ij which is also an interval j=1 ∞ P I, does |I| = |Ij|? This is guaranteed by Theorem 1.3. In fact, Theorem 1.3 can be j=1 re-phrased as follows: Theorem 1.30 Let φ be the characteristic function of a finite union of bounded intervals. n R Suppose that φn+1(x) ≤ φn(x) for all x ∈ R, and φn(x) → 0 a.e. then φn → 0. Proof Let φn = χEn , thus En+1 ⊆ En. We may assume without loss of generality that E1 is an interval. Set Fn = En \ En+1 and I = E1. Then φn(x) → 0 a.e. is the same as saying ∞ ∞ ∞ ∞ S P P P I \ Fj is null.